Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

AP Statistics

AP Statistics Practice Test: Practice Test 93

Practice Test 93 for AP Statistics: real questions and explanations from the Varsity Tutors practice-test pool.

0 / 25 answered

Question 1 of 25

A university reports that the population mean time to walk between two common campus locations is $\mu=12$ minutes, with population standard deviation $\sigma=4$ minutes. The distribution of individual walking times is somewhat skewed. A student repeatedly selects simple random samples of $n=64$ students and computes the sample mean walking time $\bar{x}$ . Which statement about the sampling distribution of $\bar{x}$ is correct?

Question Navigator

All questions

Question 1

The sampling distribution of $\bar{x}$ has mean $12$ and standard deviation $4$ , and it has the same shape as the population.
The sampling distribution of $\bar{x}$ has mean $12$ and standard deviation $4/\sqrt{64}$ , and it is approximately normal. (correct answer)
The sampling distribution of $\bar{x}$ has mean $64$ and standard deviation $4/\sqrt{12}$ .
The sampling distribution of $\bar{x}$ has mean $12/64$ and standard deviation $4/\sqrt{64}$ .
The sampling distribution of $\bar{x}$ has standard deviation $4/64$ because averaging divides by $n$ .

Explanation: This question examines the sampling distribution for walking times with μ=12 minutes and σ=4 minutes. With samples of size n=64, the sampling distribution of x̄ has mean 12 and standard deviation σ/√n = 4/√64 = 4/8 = 0.5. Since n=64 is well above 30, the Central Limit Theorem ensures the sampling distribution is approximately normal despite the somewhat skewed population. Choice A incorrectly maintains the population standard deviation, while E incorrectly divides by n=64 rather than √64.

Question 2

A website runs an A/B test on exactly 100 visitors. For each visitor, the outcome is whether they click the “Buy” button. A success is “visitor clicks,” and the site randomly assigns visitors to the same version A only (no switching). Does this meet binomial conditions to model the number of successes among the 100 visitors?

No, because the probability of a click is different for each visitor.
No, because there are three outcomes: click, no click, or leave quickly.
Yes, because there are a fixed number of trials, two outcomes per trial, approximately constant $p$ , and independence is reasonable. (correct answer)
No, because 100 visitors is too large for a binomial model.
No, because success must be a rare event.

Explanation: This question evaluates knowledge of binomial conditions in an A/B testing context, where we model clicks as successes. The conditions are met: fixed n=100 visitors, binary outcomes (click or not), approximately constant p (all assigned to version A), and reasonable independence (visitors don't influence each other). Thus, choice C is correct. Distractor A might assume varying p per visitor, but with identical conditions, p is constant; distractor B invents a third outcome, but it's binary. Choice D wrongly limits binomial to small n, but binomial works for any n. Mini-lesson: Binomial settings apply to digital experiments like this when trials are identical and independent, allowing us to calculate probabilities of k successes using the formula P(X=k) = C(n,k) p^k (1-p)^{n-k}.

Question 3

In a large city, the true proportion of commuters who take public transportation is $p=0.35$ . An analyst repeatedly selects many random samples of size $n=50$ and records the sample proportion $\hat p$ for each sample. Which statement about the sampling distribution of $\hat p$ is correct?

The sampling distribution of $\hat p$ is centered at $0.35$ , and its standard deviation decreases if $n$ increases. (correct answer)
The sampling distribution of $\hat p$ is centered at $0.35$ , and its standard deviation increases if $n$ increases.
The sampling distribution of $\hat p$ is centered at $\hat p$ , which varies from sample to sample, so it has no fixed mean.
The sampling distribution of $\hat p$ is centered at $0.35$ , and it has standard deviation $\sqrt{0.35(0.65)}$ .
The sampling distribution of $\hat p$ is centered at $0.50$ because sample proportions are equally likely to be above or below $0.50$ .

Explanation: This question assesses knowledge of how sample size affects the sampling distribution of p̂. The sampling distribution is centered at the true proportion p = 0.35, and its standard deviation σ_p̂ = √[p(1-p)/n] decreases as n increases because n is in the denominator. Choice B incorrectly states that spread increases with larger n. Choice C misunderstands that while individual p̂ values vary, the sampling distribution has a fixed mean equal to p. Choice D gives the numerator of the standard error formula but omits dividing by n. Choice E incorrectly assumes the center is at 0.50. The key concept is that larger samples produce more precise estimates with less variability.

Question 4

A streaming service defines the discrete random variable $X$ as the number of episodes a randomly selected user watches in one sitting. The probability distribution for $X$ is given in the table. Which statement correctly interprets the random variable?

$X$ is the probability that a user watches exactly 2 episodes in one sitting.
$X$ is the set of episodes available on the streaming service.
$X$ is the number of episodes a randomly selected user watches in one sitting. (correct answer)
$X$ is the probability distribution of viewing times (in minutes) for a user.
$X$ is the event that a user watches at least 3 episodes in one sitting.

Explanation: This question tests interpretation of random variables in media consumption. The random variable X counts the number of episodes a randomly selected user watches in one sitting. Choice A incorrectly interprets X as a probability rather than a count of episodes. Choice B confuses X with a set of available episodes rather than a numerical quantity. Choice D mistakes X for measuring time rather than episode count. Choice E describes X as a specific event rather than a variable that can take multiple values. When interpreting random variables, focus on what numerical quantity varies randomly - here it's the number of episodes watched, making X a discrete random variable that might take values like 1, 2, 3, etc.

Question 5

An environmental group compared the proportion of households that recycle weekly in two neighboring towns. In a random sample of 160 households in Town A, 92 reported recycling weekly; in a random sample of 150 households in Town B, 78 reported recycling weekly. A two-proportion $z$ test was conducted for $H_0: p_A=p_B$ versus $H_a: p_A\ne p_B$ , and the p-value was $0.29$ . Using $\alpha=0.05$ , what conclusion is appropriate?

Reject $H_0$ ; there is sufficient evidence that the population weekly recycling proportions differ between Town A and Town B.
Fail to reject $H_0$ ; there is not sufficient evidence that the population weekly recycling proportions differ between Town A and Town B. (correct answer)
Fail to reject $H_0$ ; therefore, Town A and Town B have exactly the same population weekly recycling proportion.
Fail to reject $H_0$ ; there is not sufficient evidence that the sample weekly recycling proportions differ between the two samples.
Reject $H_0$ ; there is sufficient evidence that Town B’s population weekly recycling proportion is greater than Town A’s.

Explanation: This problem involves a two-tailed test for comparing recycling proportions between towns. With p-value = 0.29 > α = 0.05, we fail to reject the null hypothesis. The correct conclusion is that there's not sufficient evidence that the population weekly recycling proportions differ between Town A and Town B (choice B). Choice C incorrectly claims the proportions are exactly equal - failing to reject H₀ doesn't prove equality. Choice D incorrectly refers to sample proportions. Choice E would require rejecting H₀ and having a one-tailed test. When we fail to reject H₀, we cannot conclude the null hypothesis is true; we can only say there's insufficient evidence to support the alternative hypothesis. This distinction is crucial in hypothesis testing.

Question 6

A weather station records tomorrow’s forecast category as exactly one of: sunny, cloudy, rainy, or snowy. Let event $A$ be “tomorrow is rainy” and event $B$ be “tomorrow is snowy.” The forecast system reports only one category per day, so rainy and snowy cannot both be recorded for the same day. Are the two events mutually exclusive?

No, because a day could be both rainy and snowy in real life.
Yes, because in this setup only one category is recorded, so $A\cap B$ is impossible. (correct answer)
No, because mutually exclusive events must be independent.
Yes, because $P(A)$ and $P(B)$ are each less than 1.
No, because the categories are related so overlap must exist.

Explanation: This question evaluates mutually exclusive events in AP Statistics, characterized by no possible joint occurrence, so A ∩ B = ∅. Events A (rainy) and B (snowy) are mutually exclusive because the forecast assigns only one category per day, preventing both from being recorded together. Choice B correctly states yes, as the setup makes intersection impossible. Choice A is a distractor that confuses real-world possibilities with the experiment's categorical constraints. In a mini-lesson, mutually exclusive events are disjoint, allowing P(A or B) = P(A) + P(B); this is distinct from independence, where P(A ∩ B) = P(A)P(B).

Question 7

A nonprofit compares the proportion of people who donate after receiving one of two email subject lines. From independent random samples, 72 of 600 recipients donated after Subject Line A and 54 of 600 donated after Subject Line B. A two-proportion $z$ test was conducted for $H_0: p_A-p_B=0$ versus $H_a: p_A-p_B>0$ , and the p-value was $0.041$ . At $\alpha=0.05$ , what conclusion is appropriate?

Fail to reject $H_0$ ; there is not convincing evidence that Subject Line A leads to a higher population donation proportion than Subject Line B.
Reject $H_0$ ; there is convincing evidence that Subject Line A leads to a higher population donation proportion than Subject Line B. (correct answer)
Reject $H_0$ ; there is convincing evidence that Subject Line B leads to a higher population donation proportion than Subject Line A.
Because the p-value is below 0.05, Subject Line A caused more people to donate than Subject Line B in the population.
Reject $H_0$ ; we can conclude only that 72/600 is greater than 54/600 for these samples, without making a population inference.

Explanation: This problem tests whether Subject Line A leads to a higher donation proportion than Subject Line B. The p-value (0.041) is less than α = 0.05, so we reject H₀. Since Ha: p_A - p_B > 0, we conclude there is convincing evidence that Subject Line A leads to a higher population donation proportion than Subject Line B. Choice D incorrectly uses causal language about the population rather than properly stating the statistical conclusion. Choice E incorrectly limits the conclusion to samples only, missing the point of hypothesis testing. When we reject H₀ in favor of a one-sided alternative, we make an inference about the population proportions based on the sample evidence.

Question 8

A researcher compares mean systolic blood pressure (mmHg) for adults who follow Diet A versus Diet B. The sample mean for Diet A was 121 and for Diet B was 125. A 99% confidence interval for the difference in population means, defined as $\mu_A-\mu_B$ , is $(-9, -1)$ . The researcher claims, “Diet A leads to a lower mean systolic blood pressure than Diet B.” Is the claim supported by the confidence interval?

No, because the interval does not include 0, so the difference could be 0.
Yes, because the entire interval is below 0, supporting $\mu_A<\mu_B$ . (correct answer)
No, because the interval is for $\mu_B-\mu_A$ , not $\mu_A-\mu_B$ .
Yes, because a 99% confidence interval guarantees Diet A is better for every individual.
Yes, because 99% confidence means there is a 99% probability that $\mu_A-\mu_B$ is negative.

Explanation: This question involves interpreting a confidence interval that is entirely negative. The 99% confidence interval (-9, -1) for μ_A - μ_B contains only negative values, meaning we are 99% confident that μ_A - μ_B < 0. This is equivalent to saying μ_A < μ_B, which directly supports the researcher's claim that Diet A leads to lower mean systolic blood pressure than Diet B. Students often misinterpret what it means when an interval doesn't include 0 or misunderstand probability statements about confidence intervals. When comparing means, if the confidence interval for the difference is entirely below zero, it provides strong evidence that the first population mean is less than the second.

Question 9

A school compares mean math test scores for students in a new tutoring program vs a standard program. Using two independent random samples, a 90% confidence interval for $(\mu_{tutor}-\mu_{standard})$ is $(2, 8)$ . Which interpretation is correct?

We are 90% confident that the tutoring program increases the population mean math score by between 2 and 8 points compared with the standard program. (correct answer)
90% of students in the tutoring program scored between 2 and 8 points higher than students in the standard program.
There is a 90% chance that the true difference in means is exactly 5 points, the midpoint of the interval.
Because 0 is not in the interval, the standard program has the higher population mean score.
We are 90% confident that $(\mu_{standard}-\mu_{tutor})$ is between 2 and 8 points.

Explanation: This question asks about interpreting a positive confidence interval for the difference in mean test scores. The interval (2, 8) for (μ_tutor - μ_standard) is entirely positive, meaning μ_tutor > μ_standard. This indicates the tutoring program results in higher average scores. The correct interpretation states we are 90% confident that the tutoring program increases the population mean math score by between 2 and 8 points compared with the standard program. Choice B incorrectly applies the interval to individual students rather than population means. When interpreting confidence intervals, remember they describe our uncertainty about population parameters (means), not the range of individual values or the probability of specific parameter values.

Question 10

A cereal manufacturer states that the mean weight of cereal in its boxes is 18.0 ounces. A quality-control inspector randomly selects $n=12$ boxes and finds a sample mean weight of $\bar{x}=17.8$ ounces. The inspector wants to test whether the true mean weight is less than 18.0 ounces. Which hypotheses are appropriate?

$H_0: \mu=18.0$ vs. $H_a: \mu<18.0$ (correct answer)
$H_0: \mu=17.8$ vs. $H_a: \mu<17.8$
$H_0: \mu\le 18.0$ vs. $H_a: \mu>18.0$
$H_0: \bar{x}=18.0$ vs. $H_a: \bar{x}<18.0$
$H_0: p=18.0$ vs. $H_a: p<18.0$

Explanation: This question involves setting up a left-tailed test for a population mean. The manufacturer claims μ = 18.0 ounces, which becomes our null hypothesis: H₀: μ = 18.0. The inspector wants to test if the true mean is less than 18.0 ounces, so we use a left-tailed alternative: Hₐ: μ < 18.0. Option B incorrectly uses the sample mean (17.8) in the hypotheses—we test claims about population parameters, not sample statistics. Option C has the inequality in H₀ pointing the wrong way for a 'less than' test. Option D uses x̄ instead of μ for the population parameter. Option E uses p, which is for proportions, not means. When the research question asks about 'less than,' use < in the alternative hypothesis.

Question 11

A record store manager creates a frequency table of music genres sold in a day. The manager notes that the relative frequency for the Rock genre is 0.25. Which statement must be true?

Exactly 25 Rock albums were sold that day.
Rock was the most popular genre sold that day.
One-quarter of all albums sold that day were in the Rock genre. (correct answer)
The total number of albums sold was 25.

Explanation: A relative frequency of 0.25 means that the category accounts for $1/4$ , or one-quarter, of the total observations. Without knowing the total number of albums sold, we cannot determine the exact frequency (count), so A and D are not necessarily true. We also cannot determine if Rock was the most popular genre, as another genre could have a relative frequency greater than 0.25.

Question 12

A manufacturer claims defects in its products fall into 4 categories with proportions 0.40 cosmetic, 0.30 packaging, 0.20 functional, 0.10 labeling. A quality-control team inspects a random sample of 100 defective products and performs a chi-square goodness-of-fit test of $H_0$ : the population defect-category proportions match the claim versus $H_a$ : they do not. The p-value is 0.001. The observed and expected counts are shown. What conclusion is appropriate at the $\alpha=0.05$ level?

Fail to reject $H_0$ because the p-value is 0.001; there is insufficient evidence of a difference.
Reject $H_0$ because the p-value is 0.001; there is evidence the population defect-category distribution differs from the claimed proportions. (correct answer)
Reject $H_0$ because the p-value is 0.001; therefore 0.40/0.30/0.20/0.10 must be the true distribution in the population.
Fail to reject $H_0$ because the p-value is less than 0.05; there is evidence the claimed proportions are correct.
Because the sample includes only defective products, a goodness-of-fit test cannot be used.

Explanation: This quality control problem involves a very small p-value. With p-value = 0.001 << α = 0.05, we strongly reject H₀. This provides convincing evidence that the population defect-category distribution differs from the manufacturer's claimed proportions. Choice A incorrectly fails to reject when p is extremely small. Choice C overstates the conclusion by claiming we've found the "true" distribution. Choice D has both the wrong decision and interpretation. Choice E incorrectly suggests the test is invalid for defective products. Key insight: very small p-values provide strong evidence against H₀, but rejecting H₀ doesn't tell us what the true distribution is.

Question 13

A music teacher studies whether minutes of daily instrument practice ( $x$ ) predicts performance rating ( $y$ ) at a recital. The teacher believes that more practice improves performance ratings in the population. Which hypotheses are appropriate for testing the teacher’s claim about the slope of the population regression line predicting $y$ from $x$ ?

$H_0: b=0$ vs. $H_a: b>0$
$H_0: \beta=0$ vs. $H_a: \beta>0$ (correct answer)
$H_0: r=0$ vs. $H_a: r>0$
$H_0: \rho=0$ vs. $H_a: \rho>0$
$H_0: \beta=0$ vs. $H_a: \beta<0$

Explanation: AP Statistics question on slope hypothesis. More practice improves ratings, positive β, H₀: β = 0 vs Hₐ: β > 0, choice B. Distractors A sample b, C and D correlations, E negative. Mini-lesson: Positive association uses > 0. Population focus. Tests practice's positive effect on performance.

Question 14

A manufacturing plant defines $X$ as the number of defective items found in a random sample of 200 items from one day’s production. Over many days, the distribution of $X$ has mean $\mu_X=4.8$ defects and standard deviation $\sigma_X=1.6$ defects. Which interpretation of the mean is correct?

In the long run, the average number of defects per 200-item sample is about 4.8. (correct answer)
Exactly 4.8 defects occur in every 200-item sample.
Most samples will contain exactly 4.8 defects because that is the mean.
The mean 4.8 means the sample can never have fewer than 4 defects.
The mean 4.8 means the range of defects is 4.8.

Explanation: This question tests understanding of the mean in quality control. The mean μ_X = 4.8 defects represents the long-run average number of defects found in 200-item samples across many days. Option A correctly states this interpretation - it's about the average over many samples, not any single sample. Option B incorrectly suggests exactly 4.8 defects occur each time (impossible since defects must be whole numbers), while option D wrongly claims the mean sets a lower bound. The mean of 4.8 tells us what to expect on average when we examine many such samples over time.

Question 15

Two sports teams are compared on the proportion of games they win at home. Let $p_C$ be the true home-win proportion for Team C and $p_D$ for Team D. A 95% confidence interval for $p_C-p_D$ is $(-0.12,\ -0.01)$ . A fan claims, “Team C has a higher home-win proportion than Team D.” Is the claim supported by the interval?

Yes, because the interval is negative, so $p_C>p_D$ .
No, because the interval is negative, suggesting $p_C<p_D$ . (correct answer)
Yes, because $0$ is not in the interval, so Team C and Team D are equal.
No, because $0$ is not in the interval, so there is no evidence of a difference.
Yes, because the interval does not include $0$ , which always means the first team is better.

Explanation: This question examines interpretation of a negative confidence interval. The 95% confidence interval for p_C - p_D is (-0.12, -0.01), which is entirely negative (below 0). When a confidence interval for a difference is entirely negative, it means p_C - p_D < 0, which translates to p_C < p_D. The fan claims "Team C has a higher home-win proportion than Team D," which would require p_C > p_D. However, the interval shows the opposite: Team D has a higher home-win proportion than Team C. The claim is not supported; in fact, it is contradicted by the interval. Remember that when interpreting confidence intervals for differences, negative values indicate the first quantity is smaller than the second, not larger.

Question 16

A town wants to estimate the proportion of all registered voters in the town who support a proposed bond measure. The town has an alphabetical list of all registered voters. A pollster starts at the top of the list and surveys the first 500 names. No random numbers are used. Which statement best describes the sample representativeness?

The sample is representative because the list includes all registered voters, so starting at the top is equivalent to random selection.
The sample may be biased because selecting the first 500 alphabetically is not random and could overrepresent certain last-name groups. (correct answer)
The sample is representative because 500 is large enough to overcome any lack of randomness.
The sample is representative because voters were not randomly assigned to support or oppose the measure.
The sample is unbiased because alphabetical order guarantees each voter has the same chance of being selected.

Explanation: This question tests understanding of systematic bias in non-random selection. Taking the first 500 names alphabetically is not random selection and can create bias related to surname patterns. Certain ethnic groups may have last names clustered in specific parts of the alphabet, leading to overrepresentation of some groups and underrepresentation of others. Additionally, some voter registration systems list names by registration date within alphabetical groups, potentially biasing toward longer-term residents. Random sampling requires using a random mechanism (like random numbers) to select individuals, ensuring each person has a known probability of selection. A simple random sample would involve numbering all voters and using random numbers to select 500, eliminating any systematic patterns in selection.

Question 17

A school nurse took a random sample of 50 students and measured their resting heart rates (beats per minute). A 92% confidence interval for the population mean resting heart rate was $(71, 76)$ bpm. Which interpretation is correct?

There is a 92% probability that the population mean resting heart rate is between 71 and 76 bpm.
If the nurse repeatedly took random samples of 50 students and built a 92% confidence interval each time, about 92% of those intervals would contain the true population mean resting heart rate. (correct answer)
About 92% of all students have resting heart rates between 71 and 76 bpm.
About 92% of random samples of 50 students will have a sample mean between 71 and 76 bpm.
The confidence level 92% means the interval endpoints 71 and 76 are each correct with probability 0.92.

Explanation: This question assesses the interpretation of a confidence interval for a population mean in AP Statistics. Choice B is correct, explaining that if random samples of 50 students are repeatedly taken and 92% confidence intervals built, about 92% of those intervals would contain the true population mean resting heart rate. A typical distractor is choice A, which mistakenly applies probability to the population mean being in this specific interval. In a mini-lesson on confidence intervals for means, the interval is calculated as the sample mean ± t* (standard error), where the confidence level determines the critical value. This setup ensures that over many samples, the specified percentage of intervals cover the true mean. Understanding this repeated sampling perspective prevents confusion with individual data points or sample means.

Question 18

A company’s customer-service call lengths are strongly right-skewed because a few calls last a very long time. The population distribution is not normal. Each day, a random sample of $n=50$ calls is taken and the mean call length $\bar{x}$ is computed. Over many days, the sampling distribution of $\bar{x}$ is observed to be approximately normal. Why is the sampling distribution approximately normal?

Because the population distribution is normal, so $\bar{x}$ must be normal for any $n$
Because the sample size is large enough for the Central Limit Theorem to make $\bar{x}$ approximately normal (correct answer)
Because any sample size makes the sampling distribution of $\bar{x}$ normal, regardless of the population shape
Because the sampling distribution is normal only when sampling is done without replacement
Because the standard deviation of $\bar{x}$ equals the population standard deviation $\sigma$

Explanation: This question tests understanding of the Central Limit Theorem (CLT). The problem states that call lengths are strongly right-skewed (not normal), but when we take samples of size n=50 and compute sample means, the sampling distribution of x̄ becomes approximately normal. This happens because n=50 is large enough for the CLT to apply. The CLT states that for sufficiently large sample sizes (typically n≥30), the sampling distribution of the sample mean will be approximately normal regardless of the population's shape. Option A is incorrect because the population is explicitly not normal. Option C overstates the CLT - small samples don't guarantee normality. Options D and E describe unrelated concepts.

Question 19

A basketball player takes 20 free throws during practice. For each shot, define a success as making the free throw. Assume each shot is independent and the probability of success on any one shot is $0.75$ . Let $X$ be the number of made free throws. Which values correctly model this situation with a binomial distribution (identify $n$ and $p$ )?

$n=20,\ p=0.25$
$n=0.75,\ p=20$
$n=20,\ p=0.75$ (correct answer)
$n=15,\ p=0.75$
$n=0.25,\ p=20$

Explanation: This question requires identifying binomial parameters when success is clearly defined. The basketball player takes n = 20 shots (the number of trials), and success is defined as making the free throw with probability p = 0.75. Since X counts made free throws (successes), we use p = 0.75, not 0.25. Students often confuse which probability to use - always match p to what you're counting. Notice that choices B and E incorrectly swap n and p values, which is impossible since n must be a positive integer and p must be between 0 and 1. In binomial distributions, always identify what constitutes a 'success' first, then p is the probability of that success.

Question 20

A jar contains a mix of red and blue marbles, and each marble is either large or small. One marble is selected at random. Let event $A$ be “the marble is red,” and event $B$ be “the marble is large.” The jar was filled so that the proportion of red marbles among the large marbles is smaller than the overall proportion of red marbles in the jar. How does knowing that $B$ occurred affect the likelihood of $A$ ?

It makes $A$ more likely because $P(A\mid B)=P(B\mid A)$ .
It makes $A$ less likely because $P(A\mid B)<P(A)$ . (correct answer)
It does not change the likelihood because $P(A\mid B)=P(A)$ .
It makes $A$ more likely because $P(A\mid B)>P(A)$ .
It makes $A$ less likely because $P(A\mid B)=P(A\cap B)$ .

Explanation: Addressing conditional probability in AP Statistics, this question involves marble color (A: red) given size (B: large). The lower red proportion among larges means P(A|B) < P(A), reducing likelihood. Conditioning on B confines the sample space to large marbles, with fewer reds relatively. Choice E distracts by mixing conditional with joint, but P(A|B) ≠ P(A ∩ B) generally. Mini-lesson: Conditioning adjusts P(A) downward if B selects a subgroup with lower A prevalence, using P(A|B) = P(A ∩ B)/P(B). Knowing it's large thus decreases the red probability below overall. This demonstrates negative association in probability.

Question 21

A researcher estimates that 64% of households own a pet. Many random samples of $n=10$ households are taken and $\hat p$ is recorded. Which statement about the sampling distribution is correct?

The sampling distribution of $\hat p$ may not be approximately Normal because the success-failure condition may fail. (correct answer)
The sampling distribution of $\hat p$ is approximately Normal because $p$ is greater than 0.50.
The sampling distribution of $\hat p$ is exactly Normal for any $n$ .
The sampling distribution of $\hat p$ has mean 10.
The sampling distribution of $\hat p$ has no spread because the population proportion is fixed.

Explanation: This question evaluates normality conditions for sampling distributions of sample proportions in AP Statistics. May not be normal because n=10, n p=10*0.64=6.4<10, failing condition. Center p=0.64, but shape skewed. Choice B is distractor, wrongly tying to p>0.50. In a mini-lesson, need n p ≥ 10, n(1-p) ≥ 10 for normality; here n(1-p)=3.6<10 too. Small n limits approximation. Use exact methods instead.

Question 22

A company runs an A/B test on its website and observes that Version B has a higher average purchase amount in the sample. The company plans to roll out Version B to all users, which could increase revenue but could also reduce sales if the observed difference was due to chance. Why is it important to consider potential error when interpreting the test result?

Because any observed difference in sample means must be caused by the website version and cannot be due to chance
Because statistical evidence is never certain, and acting as if the sample result is a guarantee can lead to a costly rollout decision (correct answer)
Because error only matters when the two versions have exactly equal sample means
Because randomization makes Type I and Type II errors impossible
Because potential error refers only to computer coding mistakes, not to inference from samples

Explanation: This question evaluates the role of error in A/B testing for business strategies, such as website changes. In this context, rolling out Version B could boost or harm revenue if the sample difference was chance-based. The correct answer B stresses that statistical results aren't certain, so treating them as guarantees risks costly errors. Distractor A wrongly assumes observed differences are always real in randomized tests, ignoring possible chance effects. Mini-lesson on uncertainty: Even in controlled experiments, random assignment doesn't eliminate variability; p-values and significance tests quantify the chance of error, guiding whether to act on results.

Question 23

A school counselor wants to understand how 9th-grade students at Central High prefer to receive academic support. The counselor surveys a random sample of 120 ninth graders and records each student’s preferred support method (a categorical variable). Results are summarized below.

Which statement is supported by the data in the table?

Preferred support method	Frequency	Relative frequency
One-on-one tutoring	42	0.350
Small-group sessions	30	0.250
Online resources	24	0.200
Teacher office hours	18	0.150
Peer mentoring	6	0.050

More students preferred small-group sessions than online resources.
About $35\%$ of surveyed students preferred one-on-one tutoring, the highest proportion. (correct answer)
Teacher office hours were preferred by about 18% of surveyed students.
Peer mentoring was preferred by about 6% of surveyed students.
A majority of surveyed students preferred either online resources or teacher office hours.

Explanation: This question assesses the skill of representing a categorical variable with tables by interpreting frequency and relative frequency to identify supported statements about students' preferred academic support methods. The data supports choice B because the relative frequency for one-on-one tutoring is $0.350$ , or $35\%$ , which is indeed the highest proportion among the options, as confirmed by comparing it to $0.250$ , $0.200$ , $0.150$ , and $0.050$ . Evidence from the table shows 42 students chose this out of 120, yielding exactly $35\%$ , outperforming others like small-group sessions at $25\%$ . A common distractor is choice C, which claims about 18% preferred teacher office hours, but the actual relative frequency is $0.150$ or 15%, likely confusing the frequency count of 18 with a percentage. Another distractor, choice D, states about 6% for peer mentoring, but it's precisely 5%, highlighting the need for accurate reading. In interpreting categorical tables, a key lesson is to use relative frequencies for proportions and compare them directly to identify modes or rankings, ensuring calculations align with the total sample size for validity.

Question 24

A manufacturing plant inspects 20 light bulbs from a shipment. A bulb is a success if it is defective (so it will be returned). The long-run defect rate for this supplier is $0.03$ , and the inspections are treated as independent. Which values correctly model this situation with a binomial random variable $X$ = number of successes?

$n=20,\ p=0.97$
$n=0.03,\ p=20$
$n=20,\ p=0.03$ (correct answer)
$n=17,\ p=0.03$
$n=20,\ p=0.20$

Explanation: This question tests parameter identification when "success" is defined counterintuitively. We inspect 20 bulbs (n = 20), and "success" means the bulb is defective, which happens with probability 0.03, giving p = 0.03. Choice A incorrectly uses p = 0.97, which is the probability of NOT being defective. Choice B swaps n and p values. Note that "success" in statistics doesn't mean something good—it's simply the outcome we're counting, even if it's defects or failures in the real-world context.

Question 25

A wildlife agency wants to estimate the proportion of all fish in a lake that are a protected species. Biologists set nets only near the shoreline because it is safer and easier, then identify the species of the fish caught in those nets. They use the sample proportion from the shoreline catch to estimate the lakewide proportion. Which issue most threatens the validity of this estimate?

Undercoverage because fish that live in deeper water may be less likely to be sampled (correct answer)
Nonresponse bias because some fish avoid the nets
Response bias because fish might behave differently when they see the biologists
Measurement error because species identification is impossible in the field
Voluntary response because fish choose whether to swim into the nets

Explanation: This AP Statistics question examines potential sampling problems in estimating the proportion of protected fish species in a lake by netting only near the shoreline. Undercoverage is the primary threat, as deeper-water fish are less likely to be caught, potentially underestimating or overestimating the proportion if species differ by habitat. Voluntary response (choice E) is a distractor, but fish don't 'choose'—the issue is the limited sampling area excluding parts of the population. To diagnose, evaluate if the method misses subgroups based on location or behavior. A mini-lesson: Ensure sampling covers the full environment to avoid bias in ecological estimates. This promotes accurate population proportion inferences.